I need to find an example of a metric space , in which

${lim}_{n→∞} (1/n)$ it's different from $0$.

I took the set of real numbers with the discrete metric space $(R,d)$ in which this limit does not exist , but I'm not sure if my problem is solved or i need to find a metric space where this limit exist but its different from $0$?

  • $\begingroup$ Take $\mathbb R$ with the usual metric, and rename $0$ to "banana". So now $\lim 1/n$ is "banana" and clearly banana $\ne 0$. $\endgroup$ – Ittay Weiss Dec 2 '15 at 9:19
  • $\begingroup$ You could argue that you solved the question already. Indeed, if you consider $(\mathbb{R},d_{discrete})$, this limit does not exist, so it's certainly not zero. I'm wondering whether there is a metric on $\mathbb{R}$ such that this limit exists and is not zero. $\endgroup$ – Mathematician 42 Dec 2 '15 at 9:25

take the real numbers but define $d(x,y) = |x-y|$ if $x$ and $y$ are not 0 or 1.

define $d(1,x) = |x|$ for all $x \neq 0,1$ and define $ d(0,y) = |y-1|$ for all $y\neq 0,1.$

In other words, swap 0 and 1. Define $d(0,1)=1.$

  • $\begingroup$ This is a useful type of example for many similar questions. As a note of interest, you can also think of this as the metric induced by a function that sends 0 to 1, 1 to 0, and x to itself otherwise. Indeed one can always define a metric induced by any injection $f$ by setting $d(x,y)=|f(x)=f(y)|$. $\endgroup$ – Luke Hamblin Dec 2 '15 at 10:08
  • $\begingroup$ I've written the same myself, but you raced me. :) +1 $\endgroup$ – CiaPan Dec 2 '15 at 10:09
  • $\begingroup$ Doesn't a metric space necessitate that $d(x,y)=0 \iff x=y$? In this case, $d(0,1) = 0$ but $1\neq 0$. $\endgroup$ – mathochist Dec 2 '15 at 10:25
  • $\begingroup$ I've made it more precise $\endgroup$ – Mark Joshi Dec 2 '15 at 10:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.